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Quick 'n' Dirty Review of Vector Operations
METR 3123. Alan Shapiro, Instructor
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Scalars
A quantity that has magnitude only is called a
scalar. Examples: temperature T, pressure p,
east-west speed u, gas constant Rd.
Scalars can be constant (like Rd) or functions of
space and time (like T, p, u).
Vectors
A quantity that has magnitude and direction is a
vector. e.g., acceleration a , velocity u , temp
gradient ∇T , earth's angular velocity vector Ω .
Vectors can also be constant (like Ω ) or
functions of space and time (like a , u , ∇T ).
A vector can be represented by an arrow:
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Unit Vectors
A unit vector is a vector that has a magnitude of
1 (and is dimensionless). They are useful for
expressing directions.
Consider an arbitrary vector b . Its magnitude
can be written as b ≡ b . To get the unit vector
in the direction of b , divide b by its magnitude:
b≡ b
b
The hat means "unit vector". Multiplying this
equation through by the magnitude b yields:
b = bb
This deceptively simple equation is extremely
important. It means:
"Any vector can be written as the product of
its own magnitude and the unit vector in its
own direction"
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Vector Addition
Consider 2 vectors, c and d :
c
d
Adding c to d (tip of first to tail of the second)
yields a new vector, c + d :
c
d
c + d
Get the same vector by adding d to c :
c
d
c + d
d + c
c
d
So, vector addition is commutative: c + d = d + c
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Vector addition is also associative:
a+b+c= a+b +c = a+ b+c .
a
b
c
a+b+c
a+b
c
(a + b) + c
a
b+c
a + (b + c)
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Vector Subtraction
Similar to vector addition: just add the negative
of a vector.
c – d = c + –d
–d
c
d
c–d
–d
c
d
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Scalar product (dot product)
Consider two vectors a and b . Let γ be the
smallest angle between them:
b
γ
a
The scalar (dot) product between a and b is:
a ⋅ b = a
b cos γ .
It's a scalar! Can also write it as,
a ⋅ b = ab cos γ , where a = a , b = b .
The dot product is commutative: a ⋅ b = b ⋅ a ,
and distributive: a ⋅ (b + c) = a ⋅ b + a ⋅ c .
If a ⊥ b then a ⋅ b = 0 .
If a ⋅ b = 0 then there are 3 possible scenarios:
(i) a = 0, (ii) b = 0, (iii) a ⊥ b .
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Unit vectors of Cartesian coord system
Consider a right-handed Cartesian coordinate
system with unit vectors i, j, k . i, j, k are
mutually ⊥ .
z
k
j
y
i
x
i = 1,
j = 1,
i⋅i = i
i cos 0 = 1
i⋅j = i
j cos 90 = 0
i⋅k = i
k = 1.
i
i
j
i
k
k cos 90 = 0
i
Similarly, j ⋅ i = 0 , j ⋅ j = 1 , j ⋅ k = 0 ,
k ⋅ i = 0, k ⋅ j = 0, k ⋅ k = 1
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Projection (component)
The projection (component) of a vector c in a
particular direction is the dot product of c with
the unit vector in that particular direction. It is
the "amount" of c in that particular direction.
For example, the component of c in the vertical
direction is c ⋅ k . Call it c z . So c z ≡ c ⋅ k
= c k cos γ = c cosγ , where γ is the angle
between c and k , and c ≡ c .
c cosγ
c sin γ
k
γ
c
j
i
Similarly, the components of c in the i and j
directions are: c x ≡ c ⋅ i , and c y ≡ c ⋅ j .
Can see that c = c x i + c y j + c zk , i.e., c = sum of
its components in the i, j, and k directions.
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More on addition
Component of the sum of 2 vectors is equal to
the sum of the components of the 2 vectors.
In other words, consider vectors a and b , and
their sum c ≡ a + b . Write a , b and c as,
a = a x i + a y j + a zk ,
b = b x i + b y j + b zk ,
c = c x i + c y j + c zk .
Then the sum of a and b is given by,
c = ax + b x i + ay + b y j + az + b z k
So the i-component of c is: c x = a x + b x .
So the j-component of c is: c y = a y + b y .
So the k-component of c is: c z = a z + b z .
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More on the scalar (dot) product
With a and b written as, a = a x i + a y j + a zk ,
b = b x i + b y j + b zk , the scalar product a ⋅ b
becomes :
a ⋅ b = (a x i + a y j + a zk) ⋅ (b x i + b y j + b zk)
= a x i ⋅ (b x i + b y j + b zk) +
a y j ⋅ (b x i + b y j + b zk) +
a zk ⋅ (b x i + b y j + b zk)
= a xb x i ⋅ i + a xb y i ⋅j + a xb z i ⋅k +
a yb x j ⋅ i + a yb y j ⋅j + a yb z j ⋅k +
a zb x k ⋅ i + a zb y k ⋅j + a zb z k ⋅k
Since i, j, k are of unit length and are ⊥ to each
other, i ⋅ i = 1, i ⋅ j = 0, i ⋅ k = 0 , etc. So
previous equation boils down to:
a ⋅ b = a xb x + a yb y + a zb z
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Vector product (cross product)
Consider two vectors a and b . Let γ be the
smallest angle between them:
b
γ
a
The cross product a × b is a vector with 3
properties:
(1) a × b is ⊥ to both a and b (property 2
excludes one of two possible orientations).
(2) Right-hand rule for direction of a × b :
align fingers of your right hand with a then
curl your fingers toward b . Your thumb
indicates direction of a × b .
(3) Magnitude of a × b = area of trapezoid
formed by a and b :
a × b = ab sin γ ,
where a ≡ a , and b ≡ b .
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These three properties can be summarized on
the following diagram:
a×b
b
γ
area = ab sinγ
a
The angle γ between a and b that gives the
largest magnitude of a × b is 90° (sin90 = 1,
area is maximum). But when γ = 90°, the dot
product a ⋅ b is 0 (since cos 90 = 0).
When a and b are parallel to each other (γ = 0°
or 180°, sinγ = 0, area = 0) the cross product
a × b is 0. But when γ = 0° or 180°, the dot
product a ⋅ b has its largest magnitude.
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a×a = 0
a
a
i × i = 0,
i × j = k,
i×k = – j
k
k
i
i
i
–j
j
i
Similarly,
j × i = – k,
j × j = 0,
j×k = i,
k×i = j ,
k× j = – i,
k×k = 0
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It can be shown that a × b can be written as a
determinant:
a×b =
i
j
k
ax
ay
az
bx
by
bz
= i ay b z – azb y + j azb x – ax b z
+ k ax b y – ay b x .
The cross product is not commutative because
interchanging 2 rows of a determinant changes
it's sign: a × b = – b × a . This also follows
from the right hand rule.
However, the cross product is distributive:
a × b+c = a×b + a×c.
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Scalar triple product
Consider any 3 vectors, a, b and c , and form the
expression,
a⋅ b×c .
This is a scalar known as the scalar triple
product. (Why is it a scalar? a is a vector and
b × c is a vector. The dot product between two
vectors -- in this case a and b × c -- is a scalar.)
What does a ⋅ (b × c) look like when expanded
out?
i
j
k
Recall that b × c = b x b y b z , or equivalently,
cx cy cz
b × c = i b y cz – b zcy + j b zcx – b x cz
+ k b x cy – b y cx .
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Now take the dot product of a with b × c :
a⋅ b×c
= i ax + j ay + k az ⋅ b × c .
Plug in the expression for b × c and use fact that
i ⋅ i = 1, i ⋅ j = 0, i ⋅ k = 0 , etc, to get,
a ⋅ (b × c) = a x(b yc z – b zc y) + a y(b zc x – b xc z)
+ a z (b x c y – b y c x) .
This means that, a ⋅ b × c =
ax ay az
bx by bz .
cx cy cz
If you interchange 2 rows once, the determinant
changes sign. If you interchange 2 rows twice,
the determinant stays the same. So:
a⋅ b×c = –c⋅ b×a = –a⋅ c×b
a⋅ b×c = b⋅ c×a = c⋅ a×b
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Vector triple product
The expression,
a× b×c
is a vector known as the vector triple product.
By examining the components of a × b × c it
can be shown that,
a× b×c = b a⋅c – c a⋅b .
The location of the parentheses does matter!
In general, a × b × c ≠ a × b × c .
To see this for a particular case, let b = a and
compare a × a × c with a × a × c .
a × a × c = a a ⋅ c – c a ⋅ a , while
a × a × c = 0 . Thus, they are not equal.
0
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Differentiation rules
Suppose a and b are vectors and m is a scalar.
Suppose a, b and m are all functions of a scalar
(say, time t). Then,
d a + b = da + db ,
dt
dt
dt
d ma
dt
= m da + dm a ,
dt
dt
d a ⋅ b = da ⋅ b + a ⋅ db ,
dt
dt
dt
d a × b = da × b + a × db .
dt
dt
dt
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Directional derivative. Del operator ∇
The rate of change of temperature T in the i
(east) direction is ∂T
∂x . Similarly, the rate of
change of temperature in the north and vertical
∂T , respectively.
directions are ∂T
,
and
∂y
∂z
What about the rate of change of temperature in
an arbitrary direction? Consider the direction
specified by a unit vector m . Consider a tiny
vector element δm pointing in that direction.
δm
m
k
δx
i
δz
j
δy
δm = δx i + δy j + δz k , δm = δm ,
δy
δx
δm
m=
=
i +
j + δz k .
δm
δm
δm
δm
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δm extends from (x,y,z) to (x+δx, y+δy, z+δz).
The temperature difference δT across this tiny
distance δm is:
δT = T(x+δx, y+δy, z+δz) – T(x, y, z) .
Consider Taylor expansion of T about the
starting point x, y, z:
T(x+δx, y+δy, z+δz) = T(x, y, z) +
+ ∂T (x+δx – x) + ∂T (y+δy – y)
∂x
∂y
+ ∂T (z+δz – z) + higher order terms
∂z
∴
T(x+δx, y+δy, z+δz) – T(x ,y, z) =
∂T δx + ∂T δy + ∂T δz + h.o.t
∂x
∂y
∂z
∴
δT = ∂T δx + ∂T δy + ∂T δz + h.o.t.
∂x
∂y
∂z
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To get a rate of change of temperature in the
direction of interest, divide δT by length δm.
δT = ∂T δx + ∂T δy + ∂T δz + h.o.t
∂x δm ∂y δm ∂z δm
δm
δm
Can rewrite this using dot product notation,
δT =
δm
∂T
∂x
∂T
∂y
∂T
∂z
⋅
δx
δm
δy
δm
δz
δm
+ h.o.t
δm
δy
δx
i+
j + δz k :
or, using m =
δm
δm
δm
δT =
δm
∂T
∂x
∂T
∂y
∂T
∂z
⋅ m + h.o.t
δm
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In the limit of δm → 0, h.o.t → 0 and h.o.t/δm
→ 0 since h.o.t ~ (δm)2. So get
∂T ≡ lim δT =
∂m δm → 0 δm
∂T
∂x
∂T
∂y
∂T
∂z
⋅m
Introduce the del operator ∇ defined by,
∇ ≡ i ∂ + j ∂ + k ∂
∂x
∂y
∂z
∇ acting on a scalar f is ∇f. It's the gradient of f,
a vector. Temperature gradient is given by,
∇T = i ∂T + j ∂T + k ∂T =
∂x
∂y
∂z
∂T
∂x
∂T
∂y
∂T
∂z
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So the rate of change of T in the direction m
(directional derivative in direction of m ) is given
by,
∂T = ∇T ⋅ m .
∂m
Important! If you consider a direction m to be
perpendicular to ∇Τ then ∂T ∂m = 0 , i.e.,
there is no change in that direction. In other
words, ∇Τ is ⊥ to surfaces of constant T.
Also note that the largest positive value of
∂T ∂m is attained when m is in the direction of
∇Τ. So ∇Τ points in the direction of greatest
change in T -- from lower to higher values of T.
These ideas are illustrated in this diagram:
c
c
o
∇T
o
∇T
hot
l
∇T
o
o
l
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Divergence
Take ∇ ⋅ (an arbitrary vector field a ):
∇ ⋅ a = i ∂ + j ∂ + k ∂ ⋅ ia x + ja y + ka z
∂x
∂y
∂z
=
i ∂ ⋅ i ax + j ay + k az
∂x
+ j ∂ ⋅ i ax + j ay + k az
∂y
+ k ∂ ⋅ i ax + j ay + k az
∂z
Expand out all the terms and recall that,
i ⋅ i = 1, i ⋅ j = 0, i ⋅ k = 0 , etc. Get:
∂a y
∂a x
∂a z
∇⋅a =
+
+
∂x
∂y
∂z .
This is known as the divergence of a . Note that
a is a vector and ∇ is a vector operator. But
∇ ⋅ a is a scalar (much as the dot product
between 2 vectors is a scalar).
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Curl
Consider an arbitrary vector field a . See what
happens when you take ∇ × a :
∇ × a = i ∂ + j ∂ + k ∂ × ia x + ja y + ka z
∂x
∂y
∂z
=
i ∂ × i ax + j ay + k az
∂x
+ j ∂ × i ax + j ay + k az
∂y
+ k ∂ × i ax + j ay + k az
∂z
Expand out all the terms and recall that,
i × i = 0, i × j = k, i × k = – j , etc. Get:
∂a y
∂a z
∇×a = i
–
∂y
∂z
+ j
∂a x ∂a z
–
∂z
∂x
∂a y ∂a x
+ k
–
∂x
∂y .
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∇ × a is known as the curl of a . It is a vector!
Don't want to memorize that nasty formula for
∇ × a ? No problem! The curl can also be
written as a determinant:
∇ × a = i ∂ + j ∂ + k ∂ × ia x + ja y + ka z
∂x
∂y
∂z
i
=
j
k
∂ ∂ ∂
∂x ∂y ∂z
ax ay
∂a y
∂a z
= i
–
∂y
∂z
az
∂a x ∂a z
+ j
–
∂z
∂x
∂a y ∂a x
+ k
–
∂x
∂y .
Same result as before.
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Differentiation formulas involving ∇
If a and b are vectors and f is a scalar then,
∇⋅ a+b = ∇⋅ a + ∇⋅ b,
∇× a +b = ∇× a + ∇× b,
∇ ⋅ fa = f (∇ ⋅ a) + (∇f) ⋅ a ,
∇ × fa = f (∇ × a) + (∇f) × a ,
∇ ⋅ a × b = b ⋅ (∇ × a) – a ⋅ (∇ × b) ,
∇ × a × b = a (∇ ⋅ b) + (b ⋅ ∇) a
– b (∇ ⋅ a) – (a ⋅ ∇) b ,
∇ a ⋅ b = (a ⋅ ∇) b + (b ⋅ ∇) a
+ a × (∇ × b) + b × (∇ × a) .
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Further Reading on Vector Analysis
Baxandall, P., and Liebeck, H., 1986: Vector
Calculus. Oxford University Press, 550 pp.
A detailed, rigorous and painful account of vector calculus.
Much more mathematical than the other texts on this list. Not
for the squeamish.
Hay, G. E., 1953: Vector and Tensor Analysis.
Dover, 193 pp.
A good book: well-written, clear, complete, and relatively short.
The basics on vector analysis are covered in just 3 short
chapters: I, IV and V. The remainder of the book covers
applications of vector analysis and a short section on tensors.
It's a Dover paperback = cheap!
Kreyszig, E. K., 1993: Advanced Engineering
Mathematics, 7th ed. Wiley, ~ 1000 pp.
It's a good reference for many aspects of applied math
(advanced calculus, differential equations, complex analysis,
vector analysis, etc). It is well-written and is sufficiently
complete for your vector analysis needs. Can buy it online for
next to nothing.
Schey, H. M., 1992: Div, Grad, Curl and All
That, 2nd ed. W. W. Norton, 163 pp.
I prefer the Hay book to this one but maybe you'll like this one
better. It's well-written and short, but not quite as complete as
the Hay book.